在 C 中打印整数的链表

Question

I'm trying to get an integer of any size to a linked list in C. But when I print the list, a zero is printed always after the the integer. Please note that I'm adding each digit of the integer to the head.(head has the 0th place of the integer)

#include <stdio.h>
#include <stdlib.h>

struct node
{
    int digit;
    struct node* next; 
};

void get_number(struct node** head);
int create_node(int digit, struct node** head);
void printlist(struct node* head);


int main()
{
    struct node* head1 = malloc(sizeof(struct node*));

    get_number(&head1);

    printlist(head1);

    return 0;   
}

int create_node(int digit, struct node** head)
{
    struct node* tmp = malloc(sizeof(struct node*));

    tmp -> digit = digit;
    tmp -> next = *head;

    *head = tmp;

}

void printlist(struct node* head)
{
    struct node* curr = head; 
    if(!head)
        return;

    while(curr != NULL )
    {
        printf("%d",curr -> digit);
        curr  = curr -> next;
    }
}   

void get_number(struct node** head)
{
    int k;
    char c;

    c = getchar();
    while(c != '\n' && c != ' ')
    {
        k = c - '0';
        create_node(k, head);
        c = getchar();
    }
}

when I input 123456, the output is 1234560. I tried to find solution, but couldn't. Please help

Answer 1

You are one more node than necessary when you allocate to head1 . You simply need to call the function get_number() as:

   struct node* head1 = 0;
   get_number(&head1);

which would set the next of the last element (ie the first allocate node) to 0 and rest of the logic would be fine.

You also need to correctly call malloc() and change the type of c to int (to handle EOF ) as noted in the comments. My preferred way is to allocate memory is:

TYPE *p = malloc(sizeof *p);